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2 years ago

7

The density of lead is 30.2g/cm^3.what is the value in kilograms per meter cube?

1 answer:

V125BC [204]2 years ago

3 0

Answer:
11300 kgm3

Hope this helps

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a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

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3 years ago

Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm

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<h2>~<u>Solution</u> :-</h2>

Here, the <u>moment arm</u> is defined as follows;

The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.

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3 years ago

How do fossils provide evidence that evolution has taken place

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3 years ago

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

4 0

3 years ago

A point charge of 5.0 x 10^-7 C moves to the right at 2.6 x 10^5 m/s in a magnetic field that is directed into the screen and ha

-BARSIC- [3]

The magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

<h3>What is magnetic force?</h3>

A magnetic force is the force that act in a magnetic field.

To calculate the magnetic force, we use the formula below.

Formula:

F = qvB.........Equation 1

Where:

F = magnetic force
q = point charge
v = Velocity of the the charge
B = Field strength

From the question,

Given:

q = 5.0×10⁻⁷ C
v = 2.6×10⁵ m/s
B = 1.8×10⁻² T

Substitute these values into equation 2

F = (5.0×10⁻⁷)(2.6×10⁵)(1.8×10⁻²)
F = 23.4×10⁻⁴
F = 2.34×10⁻³ N

Hence, the magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

Learn more about magnetic force here: brainly.com/question/2279150

#SPJ12

5 0

2 years ago